Q. A $100\ W$ bulb $B_{1},$ and two $60$-$W$ bulbs $B_{2}$ and $B_{3},$ are connected to a $250\ V$ source, as shown in the figure. Now $W_{1},\ W_{2}$ and $W_{3}$ are the output powers of the bulbs $B_{1},B_{2}\ $and$\ B_{3},$ respectively. Then
A) $W_{1} > W_{2} = W_{3}$
B) $W_{1} > W_{2} > W_{3}$
C) $W_{1} < W_{2} = W_{3}$
D) $W_{1} < W_{2} < W_{3}$
Explanation
(d) $P = \frac{V^{2}}{R}$ so$\ R = \frac{V^{2}}{P} \Rightarrow R_{1} = \frac{V^{2}}{100}$ and $R_{2} = R_{3} = \frac{V^{2}}{60}$
Now $W_{1} = \frac{(250)^{2}}{\left( R_{1} + R_{2} \right)^{2}}.R_{1},\ W_{2} = \frac{(250)^{2}}{\left( R_{1} + R_{2} \right)^{2}}.R_{2}$
and$\ W_{3} = \frac{(250)^{2}}{R_{3}}$
$W_{1}:W_{2}:W_{3} = 15:25:64$ or$\ W_{1} < W_{2} < W_{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.